Addexten
Addexten is a notation that is an extension of addition. Its name is a portmanteau of "add" and "extend". It is an extension of addition, and there will be many further extensions afterward. Each extension is titled by A and the number of the extension. For example, the 13th extension will be titled A-13. For a start, the original notation, A-0. THE NOTATION IS RECURSIVE. A-0 The original notation, which is addition in disguise. x@y = x+y. x@y@z and stuff like that is NOT ALLOWED in A-0. Growth rate Notation has the order type of 1. Not even an infinity yet! A-1 There is a more powerful definition of @ where it represents recursion within the previous, like Hyper-E. The definition is: a@b = a+b, and \(a_{1}@{a_2}@ \dots a_{n-1}@a_{n} = a_{1}@a_{2}@ \dots (a_{1}@a_{2}@ \dots a_{n-1})@a_{n}-1\). The only exception is the first @, or x@y. This could be found defined all sorts of ways, starting with A-2 and later. The limit is n@n@n@n... Growth rate It behaves the same as Hyper-E, making it have the same growth rate, \(f_{\omega}(n)\). This should have somewhere between \(f(2,n)\) - \(f(3,n)\) growth rate.(IS THIS CORRECT?) As an example, 10@10 = 20, and 10@10@10 solves to 10@(10@10)@9, or 10@(10@(10@10))@8, solving to 100. n@n@n is like multiplication. So n@n@n@2 is repeatING multiplication (like 10@10@10@2 = 10@10@(10@10@10)@2). n@n@n@n is repeatED multiplication, or exponentiation. Going on, it goes up to tetration, pentation etc. Limit? \(f(\omega,n)\). A-2 Now there is a new separator: @@. Behaving a little different from Hyper-E, it's \(x@y@@2 = \underbrace{x@x \dots x@x}_{y}\). You can add additional @ separators using the same rules as A-1 before the @@ separator. There is a problem. What about @@3 or @@4? This is defined as: \(x@y@@z+1 = \underbrace{x@x \dots x@x}_{y}@@z\). This is very similar to the (1) separator in BEAF. Thus, it can be argued that A-2 and A-3 is a hybrid of E# and BEAF. Things like 10@10@@10@10 can be made. But what about stuff like 10@10@@10@@10? This uses the same property. \(x@x@@\underbrace{x@x \dots x@x}_{y} = x@y@@1@@2\). And yes, it keeps going like that. The limit is n@n@@n@@n... Now that you know the definition... Growth rate Wow... The growth needs to be analyzed more closely now, since n@n@@2 has a growth rate of \(f_{\omega}(n)\), using the definition in A-1. Since n@n@@3 recurses over n@n@@2, n@n@@3 has a growth rate of \(f_{\omega\times2}(n)\). To keep up with the pattern, n@n@@n grows at \(f_{\omega^2}(n)\). Keeping up with the recursion at n@n@@1@@2, this finds its way to the order type \(\omega^2+\omega\), because of the \(\omega\times2\)-type recursion at n@n@@2. Going further, \(\omega^2+\omega\times2\) is at n@n@@2@@2. So, n@n@@n@@2 is at \(\omega^2\times2\). Going on with this reasoning, n@n@@n@@n is at \(\omega^3\). After, n@n@@n@@n@@n is at \(f_{\omega^4}(n)\) and therefore the limit of A-2 is \(f_{\omega^{\omega}}(n)\). This is the same growth rate as linear arrays in BEAF. A-3 A-3 allows for as many @'s as you want! Another thing used for denotations: \(\underbrace{@@\dots@@}_{n} = \{n\}\). A-3 makes the notation MUCH more powerful. {n} is defined as follows: \(x@y\{n\}2 = \underbrace{x@x\{n-1\}x \dots x\{n-1\}x}_{y}\). This looks like the definition for @@ in A-2. There is a gap though. I didn't even define x@y{n}z! Simple, like last time. Here's the definition: \(x@y\{n\}z+1 = \underbrace{x@x\{n-1\} \dots x\{n-1\}x}_{y}\{n\}z\). So... that is powerful enough, but nowhere near as powerful as it can be. 10@10{X}10 isn't the limit. In fact, the @ definition applies here too. 10@10{X}10 = 10@10{10@n}10. All other previous definitions, including other ones in A-3 apply as well. The limit is n@n{n{n{...}n}n}n. Growth rate Another analysis shows that n@n{3}2 is \(f(\omega^{\omega}, n)\). Okay, so each successive @ before the {n} adds 1 to the ordinal on the FGH. Thus, each addition after the {n} separator multiplies the ordinal by 2. The @'s keep going, then n@n{3}1{2}2 can add 1 to the exponent. So n@n{3}1{2}2 has order type \(\omega^{\omega+1}\). Well then n@n{3}1{3}2 would be at \(\omega^{\omega\times2}\), and n@n{4}2 at \(\omega^{\omega^2}\). n@n{5}2 is at \(\omega^{\omega^3}\), and so n@n{n}n is at \(\omega^{\omega^{\omega}}\). Using recursion, n@n{n@n}n is just adding one to the previous ordinal, but n@n{n{2}n}n is at \(\omega^{\omega^{\omega\times2}}\). AND THEN... Assume the process keeps going! The overall limit for A-3 is... \(f(\varepsilon_0,n)\)!! A-4 A-4 defines a hyperseparator ~. It is defined by iterations of {X}. The base case, which is a@b{1 ~ 2}2 = \(a@a\underbrace{\{a\{a\{\dots\}a\}a\}}_{b}a\), can't just be it. That's at \(\varepsilon_0\), which is the limit of A-3. After that you should be able to replicate the procedures before. a@a{a{a{a{...}a}a}a}a{1 ~ 2} is a@b{1 ~ 2}3 and continuing after that... Of course there needs to be a definition after {1 ~ 2}, so it acts like before: a@b{c+1 ~ 2}2 = \(a@a\underbrace{\{c ~ 2\}a...a\{c ~ 2\}a}_{b}\). Keep on like before and eventually... a@a{a{a{a{...} ~ 2} ~ 2} ~ 2}. This is a@b{1 ~ 3}. All previous rules can be generalized too: a@b{c+1 ~ d}2 = \(a@a\underbrace{\{c ~ d\}a...a\{c ~ d\}a}_{b}\) and a@a{a{a{a{...} ~ c-1} ~ c-1} ~ c-1} with b repetitions is a@b{1 ~ c}2. Again, using previous rules, you can get to the limit: a@b{1 ~ {1 ~ {...}}}2. Growth rate This will be made at certain ordinal levels. For example, something like {1~2} = \(\varepsilon_0\) will be used. It's clearly not the case, but this is what the denotation of '=' will be for now. {1 ~ 3} = \(\varepsilon_1\) {1 ~ 4} = \(\varepsilon_2\) {1 ~ 5} = \(\varepsilon_3\) {1 ~ {1 ~ 2}} = \(\varepsilon_{\varepsilon_0}\) {1 ~ {1 ~ {1 ~ 2}}} = \(\varepsilon_{\varepsilon_{\varepsilon_0}}\) LIMIT = \(\zeta_0\). Category:RECURSIVE NOTATIONS Category:RELATED TO SEPARATORS Category:SUPERPOTENTIAL FOR EXTENDING MARK